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A train approaches a tunnel PQ. Inside a tunnel there is a rat at a point, the distance of which from P is equal to 5/12th of the length of PQ. When the train whistles, the rat runs towards the Point P of the tunnel to get exit then the Train catches it at Point P and if rat moves towards the Q for the exit then the Train catches it at Point Q. The speed of the train is how many times more than the speed of the rat?

A train approaches a tunnel PQ. Inside a tunnel there is a rat at a point, the distance of which from P is equal to 5/12th of the length of PQ. When the train whistles, the rat runs towards the Point P of the tunnel to get exit then the Train catches it at Point P and if rat moves towards the Q for the exit then the Train catches it at Point Q. The speed of the train is how many times more than the speed of the rat? Correct Answer 6

Formula used:

Distance = Speed × Time

Calculation:

Let the speed of train be x and rat be y

Suppose, the train whistles at Point T which is D away from P and the rat is at point R

According to the question:

⇒ PR = 5/12 × PQ

Suppose PR = 5k and RQ = 7k

Again,

D/x = 5k/y

⇒ x/y = D/5k     --(1)

Also, (D + 12k)/x = 7k/y

⇒ x/y = (D + 12k)/7k     --(2)

From (1) and (2), we get

D/5k = (D + 12k)/7k

⇒ 7D = 5D + 60k

⇒ D = 30k

Using the value of D in (1), we get

x/y = 30k/5k = 6/1

∴ The speed of the train is 6 times more than the rat.

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